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Pixel data (ARGB) question

Posted on 2004-08-23
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Last Modified: 2008-02-01
Hi guys,

I am currently studying on how Java stores the ARGB of a pixel data into an integer. However, I have a little problem understanding one area. This is it:

According to a certain article, an integer in Java is 32 bits (4 bytes). The first 8 bits store the alpha, the next 8 bits store the red component, the next 8 bits store the green component, and the final 8 bits store the blue component. Now, imagine I have an opaque white color. Does this mean that the integer that store this pixel data can be represented this way in hex: 0xFFFFFFFF? If yes, here comes my problem. How can 0xFFFFFFFF fit into one integer? This value exceeds the limit of an integer value. If I remember correctly, the first bit is always the sign type of the number (0 means positive, 1 means negative). Yet the article claims that "FF" is used for opaque (no transparency) colors.

What is happening?
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Question by:dr0zaxx
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12 Comments
 
LVL 14

Expert Comment

by:Farzad Akbarnejad
ID: 11878548
Hi,
We have Signed and Unsigned Integer. In Unsigned Integer we haven't sign bit. Unsigned 32 bit Integers start from 0 to 4294967295.

-FA
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LVL 14

Expert Comment

by:sudhakar_koundinya
ID: 11878600
http://java.sun.com/j2se/1.4.2/docs/api/java/lang/Integer.html#toHexString(int)

This method says you can represent the value as unsigned Hex values
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LVL 92

Accepted Solution

by:
objects earned 2000 total points
ID: 11878618
> This value exceeds the limit of an integer value

No it doesn't, it uses 4 bytes which is the size of int as you stated.
How that value is interpreted as an integer is irrelevent as it uses the byte values directly.
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LVL 1

Author Comment

by:dr0zaxx
ID: 11878681
To objects:

But how come it doesn't exceed?

While it is definitely true that I am still using 4 bytes, but integers in Java are signed.

I think the limit for integer in Java is 2147483647, which is equivalent to 0x7FFFFFFF. 0xFFFFFFFF is way more than that.

Sorry but maybe you can elaborate on what you meant? I do not understand.
0
 
LVL 92

Expert Comment

by:objects
ID: 11878695
> 0xFFFFFFFF is way more than that.

because as an int that is a -ve value.
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LVL 14

Expert Comment

by:Farzad Akbarnejad
ID: 11878699
You save 4 bytes in 32 bits. The presentation of number determine its sign. If you print 0xFFFFFFFF as signed integer you will get 4294967296.
Colors don't need to be aware of sign of numbers. They need 4 bytes for storage.

-FA
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LVL 92

Expert Comment

by:objects
ID: 11878705
try this:

            int x = 0xffffffff;
            System.out.println(Integer.toHexString(x)+"="+x);
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LVL 92

Expert Comment

by:objects
ID: 11878717
> If you print 0xFFFFFFFF as signed integer you will get 4294967296.

its -1 actually :)
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LVL 1

Author Comment

by:dr0zaxx
ID: 11878738
So do you mean that it doesn't matter what the resultant integer is? I only care about what bits the integer consist of?
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LVL 92

Expert Comment

by:objects
ID: 11878758
correct
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LVL 1

Author Comment

by:dr0zaxx
ID: 11878761
Ok thanks a lot.
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LVL 92

Expert Comment

by:objects
ID: 11878784
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